Law of Sines and Cosines Notes and Worksheets - Lindsay Bowden - Free Printable

Let's solve each problem using the Law of Sines, which states:

$$
\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
$$

Where:
- $ a, b, c $ are the sides of a triangle,
- $ A, B, C $ are the angles opposite those sides respectively.

We'll go through each problem step by step.

---

1.

Given:
- Angle $ A = 53^\circ $, angle $ B = 40^\circ $, side $ b = 5 $ (opposite $ B $), find side $ x $ (side $ a $, opposite $ A $).

Step 1: Find missing angle $ C $.
Sum of angles in triangle = $ 180^\circ $:
$$
C = 180^\circ - 53^\circ - 40^\circ = 87^\circ
$$

Step 2: Use Law of Sines:
$$
\frac{x}{\sin 53^\circ} = \frac{5}{\sin 40^\circ}
$$

Solve for $ x $:
$$
x = \frac{5 \cdot \sin 53^\circ}{\sin 40^\circ}
$$

Using calculator:
- $ \sin 53^\circ \approx 0.7986 $
- $ \sin 40^\circ \approx 0.6428 $

$$
x \approx \frac{5 \cdot 0.7986}{0.6428} \approx \frac{3.993}{0.6428} \approx 6.21
$$

✔ Answer: $ x \approx 6.21 $

---

2.

Given:
- Angles $ 128^\circ $ and $ 32^\circ $, side $ c = 10 $ (opposite $ C = 32^\circ $), find side $ x $ (side $ a $, opposite $ A = 128^\circ $)

Step 1: Find missing angle $ B $:
$$
B = 180^\circ - 128^\circ - 32^\circ = 20^\circ
$$

Now use Law of Sines:
$$
\frac{x}{\sin 128^\circ} = \frac{10}{\sin 32^\circ}
$$

Solve for $ x $:
$$
x = \frac{10 \cdot \sin 128^\circ}{\sin 32^\circ}
$$

- $ \sin 128^\circ = \sin(180^\circ - 52^\circ) = \sin 52^\circ \approx 0.7880 $
- $ \sin 32^\circ \approx 0.5299 $

$$
x \approx \frac{10 \cdot 0.7880}{0.5299} \approx \frac{7.88}{0.5299} \approx 14.85
$$

✔ Answer: $ x \approx 14.85 $

---

3.

Given:
- Angles $ 99^\circ $, $ 50^\circ $, side $ b = 9 $ (opposite $ B = 50^\circ $), find side $ x $ (side $ a $, opposite $ A = 99^\circ $)

Step 1: Find missing angle $ C $:
$$
C = 180^\circ - 99^\circ - 50^\circ = 31^\circ
$$

Use Law of Sines:
$$
\frac{x}{\sin 99^\circ} = \frac{9}{\sin 50^\circ}
$$

Solve for $ x $:
$$
x = \frac{9 \cdot \sin 99^\circ}{\sin 50^\circ}
$$

- $ \sin 99^\circ \approx 0.9877 $
- $ \sin 50^\circ \approx 0.7660 $

$$
x \approx \frac{9 \cdot 0.9877}{0.7660} \approx \frac{8.8893}{0.7660} \approx 11.60
$$

✔ Answer: $ x \approx 11.60 $

---

4.

Given:
- Angles $ 62^\circ $, $ 65^\circ $, side $ c = 6 $ (opposite $ C = 65^\circ $), find side $ x $ (side $ a $, opposite $ A = 62^\circ $)

Step 1: Find missing angle $ B $:
$$
B = 180^\circ - 62^\circ - 65^\circ = 53^\circ
$$

Use Law of Sines:
$$
\frac{x}{\sin 62^\circ} = \frac{6}{\sin 65^\circ}
$$

Solve for $ x $:
$$
x = \frac{6 \cdot \sin 62^\circ}{\sin 65^\circ}
$$

- $ \sin 62^\circ \approx 0.8829 $
- $ \sin 65^\circ \approx 0.9063 $

$$
x \approx \frac{6 \cdot 0.8829}{0.9063} \approx \frac{5.2974}{0.9063} \approx 5.84
$$

✔ Answer: $ x \approx 5.84 $

---

5.

Given:
- Angles $ 110^\circ $, $ 46^\circ $, side $ b = 15 $ (opposite $ B = 46^\circ $), find side $ x $ (side $ a $, opposite $ A = 110^\circ $)

Step 1: Find missing angle $ C $:
$$
C = 180^\circ - 110^\circ - 46^\circ = 24^\circ
$$

Use Law of Sines:
$$
\frac{x}{\sin 110^\circ} = \frac{15}{\sin 46^\circ}
$$

Solve for $ x $:
$$
x = \frac{15 \cdot \sin 110^\circ}{\sin 46^\circ}
$$

- $ \sin 110^\circ = \sin(180^\circ - 70^\circ) = \sin 70^\circ \approx 0.9397 $
- $ \sin 46^\circ \approx 0.7193 $

$$
x \approx \frac{15 \cdot 0.9397}{0.7193} \approx \frac{14.0955}{0.7193} \approx 19.60
$$

✔ Answer: $ x \approx 19.60 $

---

6.

Given:
- Angles $ 81^\circ $, $ 61^\circ $, side $ a = 7 $ (opposite $ A = 81^\circ $), find side $ x $ (side $ c $, opposite $ C = 61^\circ $)

Wait — angle at bottom is $ 81^\circ $, right angle is $ 61^\circ $, so third angle:
$$
B = 180^\circ - 81^\circ - 61^\circ = 38^\circ
$$

Side $ a = 7 $ is opposite $ A = 81^\circ $, and $ x $ is opposite $ C = 61^\circ $

Use Law of Sines:
$$
\frac{7}{\sin 81^\circ} = \frac{x}{\sin 61^\circ}
$$

Solve for $ x $:
$$
x = \frac{7 \cdot \sin 61^\circ}{\sin 81^\circ}
$$

- $ \sin 61^\circ \approx 0.8746 $
- $ \sin 81^\circ \approx 0.9877 $

$$
x \approx \frac{7 \cdot 0.8746}{0.9877} \approx \frac{6.1222}{0.9877} \approx 6.20
$$

✔ Answer: $ x \approx 6.20 $

---

7. Find all missing sides and angles.

Given:
- Side $ a = 12 $, angle $ A = 41^\circ $, angle $ C = 76^\circ $

Step 1: Find missing angle $ B $:
$$
B = 180^\circ - 41^\circ - 76^\circ = 63^\circ
$$

Now use Law of Sines to find other sides.

Let’s label:
- $ a = 12 $, $ A = 41^\circ $
- $ B = 63^\circ $, find $ b $
- $ C = 76^\circ $, find $ c $

Use:
$$
\frac{12}{\sin 41^\circ} = \frac{b}{\sin 63^\circ} = \frac{c}{\sin 76^\circ}
$$

First compute common ratio:
- $ \sin 41^\circ \approx 0.6561 $
- So: $ \frac{12}{0.6561} \approx 18.28 $

Now:
- $ b = 18.28 \cdot \sin 63^\circ \approx 18.28 \cdot 0.8910 \approx 16.29 $
- $ c = 18.28 \cdot \sin 76^\circ \approx 18.28 \cdot 0.9703 \approx 17.72 $

✔ Answers:
- $ B = 63^\circ $
- $ b \approx 16.29 $
- $ c \approx 17.72 $

---

8. Find all missing sides and angles.

Given:
- Side $ c = 16 $, angle $ A = 95^\circ $, angle $ C = 48^\circ $

Step 1: Find missing angle $ B $:
$$
B = 180^\circ - 95^\circ - 48^\circ = 37^\circ
$$

Now use Law of Sines:
$$
\frac{a}{\sin 95^\circ} = \frac{b}{\sin 37^\circ} = \frac{16}{\sin 48^\circ}
$$

Compute common ratio:
- $ \sin 48^\circ \approx 0.7431 $
- So: $ \frac{16}{0.7431} \approx 21.52 $

Now:
- $ a = 21.52 \cdot \sin 95^\circ \approx 21.52 \cdot 0.9962 \approx 21.44 $
- $ b = 21.52 \cdot \sin 37^\circ \approx 21.52 \cdot 0.5878 \approx 12.60 $

✔ Answers:
- $ B = 37^\circ $
- $ a \approx 21.44 $
- $ b \approx 12.60 $

---

✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $ x \approx 6.21 $ |
| 2 | $ x \approx 14.85 $ |
| 3 | $ x \approx 11.60 $ |
| 4 | $ x \approx 5.84 $ |
| 5 | $ x \approx 19.60 $ |
| 6 | $ x \approx 6.20 $ |
| 7 | $ B = 63^\circ $, $ b \approx 16.29 $, $ c \approx 17.72 $ |
| 8 | $ B = 37^\circ $, $ a \approx 21.44 $, $ b \approx 12.60 $ |

Let me know if you'd like these rounded differently or shown with more decimal precision!